change of variables and the jacobian - academic...

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Change of Variables and theJacobianPrerequisite: Section 3.1, Introduction to DeterminantsIn this section, we show how the determinant of a matrix is used to perform achange of variables in a double or triple integral. This technique generalizes to achange of variables in higher dimensions as well. Although the prerequisite for thissection is listed as Section 3.1, we will also need the fact that jAj = jAT j fromSection 3.3.

I Substitution in One VariableThe following example serves to recall the method of integration by substitutionfrom calculus:

Example 1 To computeR 51

p3x+ 1 dx; we �rst make the substitution u = 3x+ 1: Then du =

3 dx; and soZ 5

1

p3x+ 1 dx =

1

3

Z 5

1

p3x+ 1 (3 dx) =

1

3

Z 16

4

pu du

=1

3� 23u32

��164

=2

9(16

32 � 4 32 ) = 2

9(64� 8) = 112

9:

Note the factor of 3 in du = 3 dx. This indicates that the variable u covers 3 unitsof distance for each single unit of x. (It is as if u is measured in feet, while x ismeasured in yards.) Note that the length of the x-interval is only 4 units (from 1to 5), while the length of the u-interval is 12 units (from 4 to 16). The factor of 3in the du term compensates for this change. �In Example 1, the substitution variable u is a linear function of x, and so the

change in units is constant throughout the given interval. In the next example,however, the substitution is non-linear.

Example 2 ConsiderR 21

2x(x2+1)2 dx: Let u = x2 + 1: Then du = 2x dx: The integral is then

calculated asZ 2

1

2x

(x2 + 1)2dx =

Z 5

2

du

u2= � 1

u

��52

= �15��12

�=3

10:

The factor 2x in du = 2x dx indicates that the unit conversion from x to u is notconstant. As the x-interval [1; 2] is stretched into the u-interval [2; 5], the stretchingis done unevenly. For example, at x = 1, the scaling factor 2x = 2(1) = 2, and soat this point, the length of a u unit is 2 times smaller than the length of an x unit.However, at x = 1:5, the scaling factor 2x = 2(1:5) = 3, and so at this point, a uunit is 3 times smaller than an x unit.In particular, the x-interval [1:5; 1:51] (of length 0:01) is mapped to the u-interval

[3:25; 3:2801] (having length 0:0301). That is, the u-interval is approximately 3times as long, because the scaling factor is 3 at x = 1:5. The error in using 3 asthe scaling factor in this case is 0:0001, or 0:33%: As the length of the x-intervalapproaches 0, as it would in computing Riemann sums for integrals, the percenterror in the scaling factor also approaches 0. �In general, since dudx is the rate of change of u with respect to x, its presence in the

formula du = dudx dx keeps track of the amount of stretching involved in converting

from x-coordinates to u-coordinates. Thus, dudx is the desired scaling factor for a

change of variable in single-variable integration.

Andrilli/Hecker� Elementary Linear Algebra, 4th ed.� March 15, 2010

Copyright © 2010, Elsevier Inc. All rights reserved.

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I Double IntegralsWe now consider the analogous situation using two variables.

Example 3 The area of the parallelogram P indicated in Figure 1 is given by the followingdouble integral:

Area =ZZP

1 dx dy:

Converting this double integral into an iterated integral would be tedious. However,we can compute the area of P using Theorem 3.1. The vectors w1 = [2; 1] andw2 = [�1; 1] correspond to the sides of P , and so

area of P = absolute value of

�� 2 1

�1 1

�� = j2� (�1)j = 3:Let us now examine the e¤ect of a change of variables on the area. Since the

sides of P are the vectors w1 and w2, we �rst create new variables u and v tosatisfy the equation

[x; y] = uw1 + vw2 + [1; 1] = u[2; 1] + v[�1; 1] + [1; 1];

that is, x = 2u � v + 1, y = u + v + 1. Then, (x; y) vertices correspond to (u; v)vertices as follows:

(x; y) (u; v)

(1; 1) (0; 0)

(0; 2) (0; 1)

(3; 2) (1; 0)

(2; 3) (1; 1)

Thus, in converting to the (u; v) coordinate system, the parallelogram P is mappedto the unit square S shown in Figure 2. Therefore, it follows thatZZ

S

1 du dv = area of S = 1:

Since the parallelogram P does not have area 1, we must be missing a scaling factorof the type seen in the single variable case. Note that the scaling factor must beconstant in this case, as in Example 1, because the change of coordinates involvesonly linear functions. Since the area of P = 3(area of S), the scaling factor mustbe precisely 3. �

Figure 1: The parallelogram in the (x; y) system with vertices(1; 1); (0; 2); (3; 2); (2; 3)



3

Figure 2: The square in the (u; v) system with vertices (0; 0); (0; 1); (1; 0); (1; 1)

Note in Example 3 that we can work backwards to compute the vectors w1 and

w2 from the formulas for x and y as w1 =h@x@u ;

@y@u

i; and w2 =

h@x@v ;

@y@v

i: This

will work in general for all change of variable transformations. The idea behindthis is that a unit rectangle in (u; v) coordinates is mapped to a region in (x; y)coordinates that is approximated by a parallelogram whose sides are w1 and w2;as in Figure 3. The vectors w1 and w2 are tangent to the curved boundary of theactual image of the rectangle under the transformation. But di¤erentiation, alongwith �nding the tangent direction, also measures the rate of change, and so thelengths of w1 and w2 also represent the amount of stretching taking place in eachof these directions. Hence, the scaling factor needed for the change of variable is thearea of this approximating parallelogram, which, by Theorem 3.1, is the absolute

value of

��@x@u

@y@u

@x@v

@y@v

�� :

Figure 3: Converting a rectangle in (u; v) coordinates to an approximateparallelogram in (x; y) coordinates

In Section 3.3, it is proved that for any square matrix A, jAj = jAT j. Hence we

could have also found the scaling factor as the absolute value of

��@x@u

@x@v

@y@u

@y@v

�� instead.The matrix

J =

"@x@u

@x@v

@y@u

@y@v

#

is called the Jacobian matrix of the change of coordinates function�x = x(u; v)

y = y(u; v):

We will refer to jJj as the Jacobian determinant. In general, the correct scaling



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factor to change an integralRRR

f(x; y) dx dy over a region R into (u; v) coordinates

is the absolute value of the Jacobian determinant, that is,��jJj��. Therefore, if S is

the region in (u; v) coordinates that corresponds to R, thenZZR

f(x; y) dx dy =

ZZS

f(x(u; v); y(u; v))��jJj�� du dv:

Just as in the one-variable case, the scaling factor can vary if the change ofcoordinates is nonlinear, as we will see shortly.

I Polar CoordinatesThe polar coordinate system is frequently used to represent points in a two-dimensionalspace. In polar coordinates, each point P = (x; y) in the plane is assigned apair1 of coordinates (r; �); where r is the distance from the origin to P , and �is the angle between the positive x-axis and the vector having initial point atthe origin and terminal point P (see Figure 4). In all quadrants, the transfor-mation from polar coordinates to standard (rectangular) coordinates is given by�x = r cos �

y = r sin �: We can also convert from rectangular coordinates to polar coordi-

nates using�

r2 = x2 + y2

tan � = yx (when x 6= 0) :

Figure 4 Relationship between standard coordinates and polar coordinates inQuadrants I and II

It is useful to express certain double integrals in polar coordinates if the regionof integration (and/or the function involved) has radial or angular symmetry. Inthese instances, we need to compute the determinant of the Jacobian matrix inorder to include the proper scaling factor when we change coordinates.

jJj =��@x@u

@x@v

@y@u

@y@v

�� =�� cos � �r sin �sin � r cos �

�� = r cos2 � + r sin2 � = r:1The assignment of polar coordinates to a given point (x; y) is not unique. For example,

(x; y) =�p

3; 1�in rectangular coordinates can be represented as (r; �) in polar coordinates as

(2; �6); (2; 13�

6), or (�2; 7�

6). In general,

�p3; 1

�can be expressed in polar coordinates as (r; �),

where r = �q(p3)2 + 12 = �2, and � = �

6+ k�, where k is an even integer when r is positive,

and k is an odd integer when r is negative.



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If we are careful to ensure that r � 0, the absolute value of jJj is also r; and so thisis our scaling factor. Hence,ZZ

R

f(x; y) dx dy =

ZZR�

f(x(r; �); y(r; �)) r dr d�;

where R� is the region in the polar coordinate system corresponding to R. Thenext example illustrates this geometrically.

Example 4 Consider the square S in the (r; �) (polar) coordinate system with left bottom cor-ner at (2; �6 ); width �r = 0:1; and height �� = 0:1: The image R of this square in

the (x; y) system under the polar coordinate map�x = r cos �

y = r sin �is shown in Figure

5.Now, the square S has area �r�� = (0:1)(0:1) = 0:01; and thus the area of R

is approximately equal to the product of the Jacobian determinant, r = 2, with thearea of S. Hence, the area of R � 2(0:01) = 0:02:To understand this approximation, recall that the columns of the Jacobian ma-

trix represent vectors tangent at the corner point to the curved edges of R. Whenthese vectors are scaled properly by multiplying by �r and ��, respectively, theyrepresent the sides of a parallelogram (shown in Figure 6) whose area approximatesthe area of R. (In this particular case, the dot product of the columns is zero, andso the parallelogram is a rectangle.)Finally, we compute the actual area of R for comparison purposes. The actual

area of R is ��2� (the portion of the circle involved) times the di¤erences of the areasof the circles of radii 2:1 and 2:0. Hence,

area of R =��

2�(�(2:12)� �(22)) = 0:1

2�(�(0:41)) =

0:041

2= 0:0205:

Thus, in this case, the scale factor obtained from the Jacobian induces an errorof only 0:0005, or, 2:5%. Of course, in the actual integration, both �r ! 0 and�� ! 0; which makes the percent error approach 0 as well (although we do notprove this here). �

Figure 5: Image R of polar coordinate system square S in rectangular coordinates



6

Figure 6 The parallelogram formed by the columns of the Jacobian at the point(2; �6 )

Example 5 ConsiderRRR

px2 + y2 dx dy over the region R given by 0 � r � 1 + cos � in polar

coordinates (see Figure 7). Now,px2 + y2 = r; and soZZ

R

px2 + y2 dx dy =

ZZR

r � r dr d� =Z 2�

0

Z 1+cos �

0

r2 dr d�

=

Z 2�

0

�r3

3

��1+cos �0

d� =1

3

Z 2�

0

(1 + cos �)3 d�

=1

3

Z 2�

0

(1 + 3 cos � + 3 cos2 � + cos3 �) d�

=1

3

Z 2�

0

(3 cos � + cos3 �) d� +1

3

Z 2�

0

(1 + 3 cos2 �) d�:

An appeal to symmetry considerations (or a tedious computation) shows the �rstintegral equals 0. Using a double-angle formula on the second integral, we obtain

1

3

Z 2�

0

�1 + 3

�1

2+1

2cos 2�

��d� =

�5

6� +

1

4sin 2�

��2�0

=5�

3:

�

Figure 7: The region R in polar coordinates given by 0 � r � 1 + cos �



7

I Triple IntegralsThe situation for change of variables in three dimensions is similar. When con-verting an integral in (x; y; z) coordinates to an integral in (u; v; w) coordinates,any rectangular solid based at the point (x; y; z) and having sides �x; �y; and �zis mapped to a region approximated by a parallelepiped. The sides of this paral-lelepiped are the columns of the Jacobian matrix evaluated at (x; y; z) multipliedby �x; �y; and �z; respectively. Thus, by Theorem 3.1, the absolute value of theJacobian determinant

jJj =

��@x@u

@x@v

@x@w

@y@u

@y@v

@y@w

@z@u

@z@v

@z@w

��provides the correct scaling factor for converting from xyz-space to uvw-space.

That is, dx dy dz =��jJj�� du dv dw:

I Spherical CoordinatesOne coordinate system frequently used in three dimensions is spherical coordinates.If P = (x; y; z) is a point in the rectangular coordinate system and v is a vector fromthe origin to P , then P is assigned coordinates (�; �; �) in spherical coordinates,where � = jjvjj; � is the angle between the vector [0; 0; 1] and v, and � is the anglebetween the vector [1; 0; 0] and the projection of v onto the xy-plane (see Figure8). From elementary trigonometry, we �nd that

x = � sin� cos � �2 = x2 + y2 + z2

y = � sin� sin � tan � = yx ; when x 6= 0

z = � cos� cos� = zpx2+y2+z2

:

Figure 8: Spherical coordinates for P = (x; y; z)



8

Hence,

jJj =

��@x@�

@x@�

@x@�

@y@�

@y@�

@y@�

@z@�

@z@�

@z@�

�� =��sin� cos � � cos� cos � �� sin� sin �sin� sin � � cos� sin � � sin� cos �

cos� �� sin� 0

��= cos�

�� cos� cos � �� sin� sin �� cos� sin � � sin� cos �

�� (�� sin�) �� sin� cos � �� sin� sin �sin� sin � � sin� cos �

��= cos�(�2 cos� sin� cos2 � + �2 cos� sin� sin2 �)

+ � sin�(� sin2 � cos2 � + � sin2 � sin2 �)

= �2 cos2 � sin�(cos2 � + sin2 �) + �2 sin3 �(cos2 � + sin2 �)

= �2 cos2 � sin�+ �2 sin3 �

= �2 sin�(cos2 �+ sin2 �)

= �2 sin�:

Since 0 � � � � in spherical coordinates, the quantity �2 sin� is always nonnega-tive. Hence, when converting an integral from xyz-coordinates to ��-coordinates,we have

dx dy dz = �2 sin�d� d� d�:

Example 6 We �nd the volume of the region R bounded below by the upper half of the conez2 = x2 + y2 and bounded above by the sphere x2 + y2 + z2 = 8 (see Figure 9).Now,

volume of R =ZZZR

1 dx dy dz:

Converting to spherical coordinates, we have

volume of R =ZZZR

�2 sin�d� d� d�:

Since the radius of the sphere isp8; � ranges from 0 to

p8: The sides of the cone

are at a 45� angle from the z-axis, and so � ranges from 0 to �4 : Hence, changing

to an iterated integral, we obtain

volume of R =

Z 2�

0

Z �4

0

Z p8

0

�2 sin�d� d� d�

=

Z 2�

0

Z �4

0

��3

3sin�

��p8

0

d� d�

=

Z 2�

0

Z �4

0

8p8

3sin�d� d�

= �8p8

3

Z 2�

0

(cos�)

��40

d�

= �8p8

3

Z 2�

0

p2

2� 1!d�

= �8p8

3

p2

2� 1!(2�)

=32�

3(p2� 1):

�



9

Figure 9: Region R bounded below by the upper half of the cone z2 = x2 + y2 andbounded above by the sphere x2 + y2 + z2 = 8

I Cylindrical CoordinatesAnother frequently used three-dimensional coordinate system is cylindrical coordi-nates, (r; �; z); in which the r and � variables provide a polar coordinate system inthe xy-plane, and z is unchanged from rectangular coordinates. Thus,

x = r cos �

y = r sin �

z = z

:

In Exercise 3, you are asked to show that the Jacobian determinant for a transfor-mation from rectangular to cylindrical coordinates is r, and hence

dx dy dz = r dr d� dz:

I Higher DimensionsThe method we have shown for changing variables in double and triple integralsalso works in general for multiple integrals in Rn. In particular, to change fromx1x2 : : : xn coordinates to u1u2 : : : un coordinates, we must calculate the absolutevalue of the determinant of the Jacobian matrix,

jJj =

��

@x1@u1

@x1@u2

� � � @x1@un

@x2@u1

@x2@u2

� � � @x2@un

......

. . . � � �@xn@u1

@xn@u2

� � � @xn@un

��;

and then we have,

dx1 dx2 : : : dxn =��jJj�� du1 du2 : : : dun:

I New Vocabularycylindrical coordinatesJacobian determinantJacobian matrixpolar coordinatesspherical coordinates



10

I Highlights

� For the change of coordinates function�x = x(u; v)

y = y(u; v); the Jacobian matrix is

J =

"@x@u

@x@v

@y@u

@y@v

#; and its determinant, jJj; is called the Jacobian determinant:

� The scaling factor involved when converting a double integral from one setof coordinates to another is the absolute value of the Jacobian determinant.That is, if f is a function of variables x and y, R is a region in (x; y) coordi-nates, and S is the corresponding region in (u; v) coordinates, thenZZ

R

f(x; y) dx dy =

ZZS

f(x(u; v); y(u; v))��jJj�� du dv:

� When converting from (x; y) coordinates to (u; v) coordinates, dx dy =��jJj�� du dv:

In particular, in polar coordinates, where x = r cos � and y = r sin �; we havedx dy = r dr d�:

� When converting an integral in (x; y; z) coordinates to an integral in (u; v; w)coordinates, the absolute value of the Jacobian determinant

jJj =

��@x@u

@x@v

@x@w

@y@u

@y@v

@y@w

@z@u

@z@v

@z@w

��provides the correct scaling factor for converting from xyz-space to uvw-space.

That is, dx dy dz =��jJj�� du dv dw:

� In spherical coordinates, where x = � sin� cos �; y = � sin� sin �; z = � cos�;we have dx dy dz = �2 sin�d� d� d�:

� In cylindrical coordinates, where x = r cos �; y = r sin �; z = z, we havedx dy dz = r dr d� dz:

� When converting from x1x2 : : : xn coordinates to u1u2 : : : un coordinates, theJacobian matrix is

J =

2666664@x1@u1

@x1@u2

� � � @x1@un

@x2@u1

@x2@u2

� � � @x2@un

......

. . . � � �@xn@u1

@xn@u2

� � � @xn@un

3777775 ;

and we have dx1 dx2 : : : dxn =��jJj�� du1 du2 : : : dun:

I EXERCISES

1. For each change of variable formula, compute dx dy in terms of du dv.

a)F x = u+ v; y = u� vb) x = u2 + v2; y = u2 � v2

c)F x = u2 � v2; y = 2uvd) x = u

u2+v2 ; y =�v

u2+v2



11

e)F x = 2u(u+1)2+v2 ; y =

1�(u2+v2)(u+1)2+v2

2. For each change of variable formula, compute dx dy dz in terms of du dv dw:

a)F x = u+ v; y = v + w; z = w + u

b) x = 3u+ v + w; y = 3v + w; z = w

c)F x = uu2+v2+w2 ; y =

vu2+v2+w2 ; z =

wu2+v2+w2

d) x = wu ; y = u, z = u cos v (for u > 0)

3. Show that jJj = r for the change of variables from rectangular coordinates tocylindrical coordinates.

4. Compute each of the following integrals by changing to the indicated coordi-nate system:

a)FRRR

(x+ y) dx dy; where R is the region in the �rst quadrant between the

circles x2 + y2 = 1 and x2 + y2 = 9; polar coordinates

b)RRR

1 dx dy; where R is the region inside the innermost ring of the spiral

r = � in the �rst quadrant (see Figure 10); polar coordinates

Figure 10: The spiral r = �

c)FRRRR

z dx dy dz; where R is the half of the sphere of radius 1 centered at

the origin which is above the xy-plane; spherical coordinates

d)RRRR

1x2+y2+z2 dx dy dz; where R is the shell between the spheres of radii

2 and 3 centered at the origin; spherical coordinates

e)FRRRR

(x2 + y2 + z2) dx dy dz; where R is the region de�ned by x2 + y2 � 4

and �3 � z � 5; cylindrical coordinates

5.F True or False:

a) A linear change of coordinates for an integration results in a constantscaling factor with respect to the associated integrals.

b) For the change of variables u = y, v = x, we have du dv = 1 dx dy.

c) A rectangle in uv-coordinates with sides �u and �v is mapped by achange of coordinates to a region whose area is approximated by the

area of the parallelogram with sidesh@x@u ;

@y@u

i�u and

h@x@v ;

@y@v

i�v:

d) The scaling factor for a change of variables in integrals is always thedeterminant of the Jacobian matrix.



12

I Answers to Selected Exercises(1) (a) dx dy = 2 du dv

(c) dx dy = 4(u2 + v2) du dv

(e) dx dy =�

8jvj((u+1)2+v2)3

�du dv

(2) (a) dx dy dz = 2 du dv dw

(c) dx dy dz =�

1(u2+v2+w2)3

�du dv dw

(4) (a) 523

(c) �4

(e) 8003 �

(5) (a) T

(b) T

(c) T

(d) F



change of variables and the jacobian - academic...

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